A segment $AB$ is given and it's midpoint $K$. On the perpendicular line to $AB$, passing through $K$ a point $C$, different from $K$ is chosen. Let $N$ be the intersection of $AC$ and the line passing through $B$ and the midpoint of $CK$. Let $U$ be the intersection point of $AB$ and the line passing through $C$ and $L$, the midpoint of $BN$. Prove that the ratio of the areas of the triangles $CNL$ and $BUL$, is independent of the choice of the point $C$.
Problem
Source: Macedonia National Olympiad 2016
Tags: geometry
ThE-dArK-lOrD
09.04.2016 19:49
Is it real national olympiad ? Let $M$ is midpoint of $CK$ We apply a bunch of Menelaus get $AK:KU:UB=2:1:1$, $UL:LC=1:2$, $AN:NC=2:1$ and $NM:ML:LB=1:1:2$ So we can find $\frac{[ULB]}{[BMK]}$ and $\frac{[AUC]}{[CNL]}$ are fixed constant And $\frac{[BMK]}{[AUC]} =\frac{1}{2} \cdot \frac{[BMK]}{[AMU]} =\frac{1}{2} \cdot \frac{2}{3}$ which is also a fixed constant
Stefan4024
09.04.2016 21:37
ThE-dArK-lOrD wrote: Is it real national olympiad ? Let $M$ is midpoint of $CK$ We apply a bunch of Menelaus get $AK:KU:UB=2:1:1$, $UL:LC=1:2$, $AN:NC=2:1$ and $NM:ML:LB=1:1:2$ So we can find $\frac{[ULB]}{[BMK]}$ and $\frac{[AUC]}{[CNL]}$ are fixed constant And $\frac{[BMK]}{[AUC]} =\frac{1}{2} \cdot \frac{[BMK]}{[AMU]} =\frac{1}{2} \cdot \frac{2}{3}$ which is also a fixed constant Unfortunately, yes. Contestants are getting easily over 30 out of 40. The problem is too trivial. $K$ can be any point on $AB$, as long as it is fixed. Also $C$ can be any point in the plane. If we let $\frac{AK}{KB} = x$, then $\frac{[BMK]}{[AUC]} = \frac{x+3}{x+1}$.
kjytay
19.06.2016 08:15
This problem falls easily to coordinate geometry as well. (For simpler calculations, let $A = (-1, 0)$, $B=(1, 0)$, $C=(0, 2c)$.)
StefanS
25.12.2017 18:30
ThE-dArK-lOrD wrote: Is it real national olympiad ? Let $M$ is midpoint of $CK$ We apply a bunch of Menelaus get $AK:KU:UB=2:1:1$, $UL:LC=1:2$, $AN:NC=2:1$ and $NM:ML:LB=1:1:2$ So we can find $\frac{[ULB]}{[BMK]}$ and $\frac{[AUC]}{[CNL]}$ are fixed constant And $\frac{[BMK]}{[AUC]} =\frac{1}{2} \cdot \frac{[BMK]}{[AMU]} =\frac{1}{2} \cdot \frac{2}{3}$ which is also a fixed constant $\frac{[BUL]}{[CNL]}=\frac{UL}{LC}=\frac{1}{2}$
Kamran011
28.09.2019 18:50
After some Menelaus it's : $NA=2CN$ $UA=3UB$ $LC=2LU$ $UK=UB=\frac12{AK}$ Note that by Thales $CU\parallel{KN}$ Hence $\frac{[LUK]}{[CNL]}=\frac{[BUL]}{[CNL]}=\frac12$ .